Let $a,b,c\in[1,2]$ such that $a,b$ are constants then prove :
$$f(c)=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}\geq h(c)=(c-1)\frac{g(2)-g(1)}{2-1}+g(1)\geq g(c)=\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2}{ab+bc+ca}}$$
Yes it's a probable refinement of HN_NH's inequality/ Stronger than Nesbitt inequality
My refinement is based on two observations :
- The function $g(c)$ is convex on $[1,2]$
2.The chord of a convex function is greater than the convex function.
To know if the LHS is good I have tried derivatives . It gives a quartic and it's very ugly so I can say that I have not the solution for the LHS but it seems to be true (numerical check).
Update 12/11/2020:
The function :
$$p(c)=f(c)-h(c)$$
Is convex on $[1,2]$ so there is a possibility to use Jensen's inequality but now I don't see any good issue .
Using Jensen's inequality we have :
$$p(c)+p(1)\geq 2p\left(\frac{1+c}{2}\right)$$
And :
$$p\left(\frac{1+c}{2}\right)+p(1)\geq 2p\left(\frac{3+c}{4}\right)$$
And :
$$p\left(\frac{3+c}{4}\right)+p(1)\geq 2p\left(\frac{3+c}{8}+\frac{1}{2}\right)$$
And so on...Playing with these inequalities we got the result I think !
Have you an idea to show the LHS(or confirm my update)?
Thanks in advance
Max.
